The $n$th term of the series $1+3+7+13+21+\ldots$ is $9901$. The value of $n$ is

The sequence of natural numbers is divided into groups as follows: $(1), (2, 3), (4, 5, 6), (7, 8, 9, 10), \dots$. Find the sum of the numbers in the $n^{th}$ group.

$A$ person wants to climb a $n$-step staircase using one step or two steps at a time. Let $C_n$ denote the number of ways of climbing the $n$-step staircase. Then $C_{18} + C_{19}$ equals

If $\alpha \in R, n \in N$ and $n+2(n-1)+3(n-2)+\ldots+(n-1)2+n.1 = \alpha n(n+1)(n+2)$,then $\alpha =$

The sum of the infinite series $(\frac{1}{3}+\frac{4}{7})+(\frac{1}{3^{2}}+\frac{1}{3}\times\frac{4}{7}+\frac{4^{2}}{7^{2}})+(\frac{1}{3^{3}}+\frac{1}{3^{2}}\times\frac{4}{7}+\frac{1}{3}\times\frac{4^{2}}{7^{2}}+\frac{4^{3}}{7^{3}}) + \dots$ is equal to -

$\sum_{k=1}^{\infty} \sum_{r=0}^k \frac{1}{3^k} \binom{k}{r}$ is equal to